Ch. 17
PLANE MOTION OF RIGID BODIES
ENERGY AND MOMENTUM METHOD
Prof. SangJoon Shin
Active Aeroelasticity and Rotorcraft Lab.
17.0 Introduction
Method of work and energy
Method of impulse and momentum
⚫
⚫
⚫
2
Plane motion of the rigid bodies
Method of work and energy
- Work of a force and a couple
- Kinetic energy of a rigid body in plane motion
Problems involving displacement and velocities
Principle of conservation of energy
Principle of impulse and momentum
- Problem involving velocities and time
- conservation of angular momentum
Eccentric impact of rigid bodies
- coefficient of restitution
Colliding bodies moving freely
Colliding bodies partially constrained
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1A Principle of Work and Energy
for a Rigid Body
Work of a force
⚫
Main advantage
⚫
Assumption------ rigid body is made of a large number n of particles of mass mi
Kinetic energy of a particle
Scalar quantities
(17.1)
T1 + U1→2 = T2
1 n
T =  mi vi2
2 i
Positive scalar quantities
U1→2 : work of all the forces acting on the various particles of the body
(internal + external)
Total work of the internal forces =0
[Example] two particle A,B F and −F (Fig. 17.1)
displacement dr , dr ' different, but the
component along AB must be equal.
total work of the internal forces =0
U1→2 reduces to the work of the
external forces only
3
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1B Work of Forces Acting on a
Rigid Body
A2
U1→2 =  F dr
(17.3)
A1
s2
=  ( F cos  )ds
(17.3’)
s1
 : the angle which the force forms with direction of motion
s : variable of integration, which measures the distance traveled by A
Work of a couple
…….. don’t need to consider separately the work of each of the two forces forming
the couple
•
Fig. 17.2 ….two forces
F and −F forming a couple of moment M small displacement
A → A' B → B
1. A, B undergo equal displacement dr
,
1
……..sum of the work of F and −F is zero
Two part
2. A’ remains fixed. B → B ", dr2 , ds2 = rd
……only F works,
dU = Md
2
4
U1→2 =  Md
1
dU = Fds2 = Frd
M
(17.4)
(17.5)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1B Work of Forces Acting on a
Rigid Body
When M is constant, U1→2 = M ( 2 − 1 )
(17.6)
Forces applied to a fixed point
No work……..
Acting in a direction perpendicular to the displacement
[Ex] -
Reaction at a frictionless pin
Reaction of a frictionless surface
Weight of a body when its c.g. moves horizontally
+ a rigid body rolls without sliding a friction force does no work
(
velocity vc of the point contract C is zero )
dU = Fds = F (vc dt ) = 0
5
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1C Kinetic Energy of a Rigid Body
in Plane Motion
v of the mass center G
vi sum of
•
vi relative to Gx ' y ' attached at G
Kinetic energy of the rigid body
1
1 n
2
T = mv +  mi vi2
2
2 i =1
vi = ri 
(17.7)
1
1 n 2
2
T = mv + ( ri mi ) 2
2
2 i =1
I
1
1
= mv 2 + I  2
2
2
1
mv 2
- translation (  = 0 )
2
1
I 2
centroidal rotation ( v = 0 )
2
(17.8)
(17.9)
1
mv 2 associated with the motion of G
2
1
2.
I  2 associated with rotation about G
2
1.
- Rigid body in general plane motion
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1C Kinetic Energy of a Rigid Body
in Plane Motion
•
Noncentroidal Rotation
2
1 n
1 n
1 n 2

2
T =  mi ( vi ) =  mi ( ri ) =   ri mi   2
2 i =1
2 i =1
2  i =1

I0
T=
1
I 0 2
2
(17.10)
-→ applicable only in noncentroidal rotation,
prefer to use Eq.(17.9)
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1D Systems of Rigid Bodies
•
Add all the kinetic energies of all the particles, and al the forces involved
T1 + U1→2 = T2
(17.11)
U1→2 : all the forces (internal + external)
Pin-connected members
•
Problems involving
Blocks and pulleys connected by inextensible cords
Meshed gears
work of the internal forces is zero, U1→2 reduces to work of the external force
(
forces in each pair move through equal distance)
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1E Conservation of Energy
•
Work of a conservative forces
A change in potential energy
[Example] – weigh of a body, force exerted by a spring
•
Principle of work and energy
Modified form
T1 + V1 = T2 + V2
(17.12)
---sum of the kinetic energy and the potential energy of the system remains constant
1
1
2
2
(k.e = transition term mv , rotational term I  )
2
2
[Example] slender rod AB
length L, mass m, extremeties
connected blocks of negligible
mass sliding along horizontal and
vertical tracks released with no
initial velocity from a horizontal
position (Fig. 17.5 (a))
Angular velocity after rotating 
(Fig 17.5 (b))?
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1D Conservation of Energy
[sol]
T1 = 0 V1 = 0
After rotating  , G is at
1
l sin  below the reference level,
2
1
1
V2 = − Wl sin  = − mgl sin 
2
2
At this instant, instantaneous center of rotation at C,
1
1
CG = l , v2 = l
2
2
2
1 ml 2 2
1
1
1 1  1 1
2
2 2

T2 = mv2 + I 2 = m  l  +  ml  2 =
2 3
2
2
2  2  2  12

T1 + V1 = T2 + V2
1 ml 2 2 1
0=
 − mgl sin 
2 3
2
1/2
 3g

 =  sin  
 l

10
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1D Conservation of Energy
•
To determine reactions at fixed axles, rollers, or sliding blocks,
supplemented by Newton’s 2nd law
the method of work and energy
Coupled analysis ………combined use of
The principle of equivalence of the
external forces/moments and inertial
terms
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.1E Power
Power =
(13.13)
Rigid body rotating at  and acted upon by a couple of moment M
•
Power =
•
12
dV
= F v
dt
dV Md
=
= M
dt
dt
(17.13)
The various units used to measure power, such as the watt and the horsepower,
were defined in sec.13.1D
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2A Principle of Impulse and Momentum
for the Plane Motion of a Rigid Body
Principle of impulse and momentum………well adapted to the problems involving
time and velocity
•
- only practicable method for impulsive motion or impact
•
Section 14.1C …… the system of the momenta of the particle, at t1
+ the system of the impulses of the external forces at t1 ~ t2
equipollent
For rigid body,
system of the momenta at t2
equivalent
System Momenta1 + System External Impulse1->2 = System Momenta2
(17.14)
where,
n
n
i =1
i =1
L =  vi mi = mv ; H G =  ri  vi mi = I 
……. system of the momenta vi mi
13
equivalent
linear momentum mv
vectors attached at G
(Fig. 17.7)
Angular momentum
couple I 
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2A Principle of Impulse and Momentum
for the Plane Motion of a Rigid Body
Fig. 17.6 (a),(c )
Fig. 17.8
……. Impulse-momentum diagram: visual representation of Eq. (17.14)
•
Three eqns of motion from Fig. 17.8
two…..summing and equating the x and y component
third…..summing and equating the moments about any given point
(coordinate is either fixed or translated with G)
Keep the same position
relative to the
coordinate axes during
the interval of time
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2A Principle of Impulse and Momentum
for the Plane Motion of a Rigid Body
•
Sum of the moments about an arbitrary point P
I 1 + mv1d 1 +   M P dt = I 2 + mv2 d 2
t2
(17.14’)
t1
d  : perpendicular distance from P to the line of action of
linear velocity of G
•
Sum of the moments about C.G. of the body
I 1 +   M G dt = I 2
t2
t1
15
•
Careful about avoid adding linear and angular momenta
indiscriminately
•
I  should be added only to the moment of mv
(17.14”)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2A Principle of Impulse and Momentum
for the Plane Motion of a Rigid Body
•
Noncentroidal Rotation
v = r  , mv = mr  ,
- Moments about O (Fig 17.9)
H 0 = I  + (mr  )r = ( I + mr 2 ) = I o
(17.15)
- Moments about O of the momenta and impulses in
Eq.(17.14)
t2
I o1 +   M o dt = I o2
t1
(17.16)
can be used w.r.t the instantaneous axis of rotation under certain conditions
All problems of plane motion should be solved by the general method described
earlier
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2B Systems of Rigid Bodies
•
Apply the principle of impulse and moment to each body separately (Sample
Prob. 17.7)
•
No more than three unknowns
Apply the principle to the system as a whole
- Impulses of internal forces can be omitted
- each equation should be checked to make sure that consistent units have
been used.(Sampled Prob. 17.9 - 17.13)
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2C Conservation of Angular
Momentum
•
No external force
System of momenta at t1
equipollent
System of momenta at t2
total linear momentum is conserved at any direction
total angular momentum is conserved
linear momentum is NOT conserved
•
Many engineering application
Angular momentum conserved
(H ) = (H )
P
1
P
(17.17)
2
lines of action of all external forces pass….through P
sum of angular impulses of the external forces about P is zero
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.2C Conservation of Angular
Momentum
Problems involving the conservation of angular momentum about a point P
- drawing impulse-momentum diagrams as described earlier
- obtain Eq. (17.17) by summing and equating moments about P (Sample
Prob. 17.9)
•
•
19
(Sample Prob. 17.11) obtain two additional eqns. by summing and equating
x and y components of the linear momentum. Then use those to determine
two unknown linear impulses.
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.3 Eccentric Impact
•
Central impact---- mass centers of the two colliding bodies are on the line of impact
•
Eccentric Impact of two rigid bodies
--- v A and vB before impact of the two points of contact A, B (Fig. 17.10 (a))
- Period of deformation ------ at its end, v A and vB will have equal components long
the line of impact (Fig. 17.10 (b))
- Period of restitution -----at its end, v A ' and vB ' (Fig. 17.10 (c ))
•
Coefficient of restitution
Sec 13.4……..relative
velocities along the line of
impact
Rdt

e=
 Pdt
(17.18)
( vB ')n − ( vA ')n = e[( vA )n − ( vB )n ]
(17.19)
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.3 Eccentric Impact
•
For rigid body only impulsive force exerted during the impact are applied at A, B
Fig. 17.11 ….. Momentum and impulse diagram for the body A (period of deformation)
v , u ………. Velocity of the
mas
center
at
the
beginning and end of
period of deformation
 ,   ………. Angular velocities
- Components of the
momenta along the line of
impact nn
mvn −  Pdt = mun
(17.20)
-moments about G
I  − r  Pdt =I  
(17.21)
r :perpendicular distance from G to the line of impact
•
Period of restitution
mvn −  Rdt = mvn '
(17.22)
I   − r  Rdt =I  '
(17.23)
v ',  ' velocity of G, angular velocity after impact
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Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.3 Eccentric Impact
(17.20),(17.22)
(17.18)
(17.21),(17.23)
(17.18)
 −  '
e=
r
 − 
u −v '
e= n n
vn − un
(17.24)
(+)
un + r  − (vn '+ r ')
e=
un + r − (un + r  )
(17.25)
vn + r = (v A ) n un + r  = (u A ) n vn '+ r ' = (v ' A ) n
(v A ) n − (v A' ) n
e=
(v A ) n − (u A ) n
•
Second body
Similar expression for e in terms of the components along nn
Of the successive velocities of B.
and eliminating these two velocities
22
(17.26)
(17.19)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.3 Eccentric Impact
•
Constrained to rotate about O ……. Compound pendulum (Fig. 17.12(a))
impulsive reaction will be exerted at O (Fig. 17.12. (b))
Apply Eq. (17.16) to the
period of
deformation
period of
restitution
I  − r  Pdt =I  
(17.27)
I   − r  Pdt =I  '
(17.28)
:perpendicular distance from the fixed point O to the line of impact
(17.27),(17.28)
(17.18)
 * −  ' r  − r ' (v A ) n − (v A' ) n
e=
=
=
 −  * r − r  (v A ) n − (u A ) n
23
Components along nn of the successive velocities
at A
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17.3 Eccentric Impact
Eq. (17.26) still holds
Eq. (17.19) remains valid when constrained to rotate about O
(17.19) in conjunction with one or several other equations obtained by the
principle of impulse and momentum (Sample Prob. 17.11, 17. 13)
24
Active Aeroelasticity and Rotorcraft Lab., Seoul National University